Question

here, w = whole number
e = belongs to
Q. If
$w \alpha = \: ( \alpha x | \: xew)$
, then w3nw5 =

option are
1) w5 2) w3
3) w8 4) w15
If don't know what I am asking then sow the attached photo. I know 4th answer is correct but why ??? please give me explaintion. ​

Answer 1

Step-by-step explanation:

Wₐ = { ax: x ∈ W}

= set of whole number divisible by a.

W₃ = { 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, ...} = set of whole number divisible by 3.

W₅ = { 0, 5, 10, 15, 20, 25, 30, 35, 40, ....} = set of whole number divisible by 5.

If we take intersection of the two set, we get

W₃∩W₅= { 0, 15, 30,......}

= set of whole number divisible by 15

= W₁₅

If we look at this problem logically, elements in W₃∩W₅ must be divisible by both 3 and 5 since it is the intersection of W₃ and W₅. And for a number to be divisible by both 3 and 5, it should be divisible by their LCM which is 15 in this case. As such elements in W₃∩W₅ should be those divisible by 15 and hence it is W₁₅.